Solve for $x$ and $y$ using elimination. ${6x+6y = 54}$ ${3x-5y = 3}$
Solution: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Multiply the bottom equation by $-2$ ${6x+6y = 54}$ $-6x+10y = -6$ Add the top and bottom equations together. $16y = 48$ $\dfrac{16y}{{16}} = \dfrac{48}{{16}}$ ${y = 3}$ Now that you know ${y = 3}$ , plug it back into $\thinspace {6x+6y = 54}\thinspace$ to find $x$ ${6x + 6}{(3)}{= 54}$ $6x+18 = 54$ $6x+18{-18} = 54{-18}$ $6x = 36$ $\dfrac{6x}{{6}} = \dfrac{36}{{6}}$ ${x = 6}$ You can also plug ${y = 3}$ into $\thinspace {3x-5y = 3}\thinspace$ and get the same answer for $x$ : ${3x - 5}{(3)}{= 3}$ ${x = 6}$